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lgbasallote

  • 2 years ago

Prove that if n is a perfect square, then n +2 is not a perfect square

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  1. lgbasallote
    • 2 years ago
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    i suppose first step would be to assume n is perfect square

  2. lgbasallote
    • 2 years ago
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    so n = m^2

  3. lgbasallote
    • 2 years ago
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    then i suppose i find n^2

  4. lgbasallote
    • 2 years ago
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    i mean n + 2

  5. lgbasallote
    • 2 years ago
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    n + 2 would be m^2 + 2

  6. lgbasallote
    • 2 years ago
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    so now... i assume m^2 + 2 is a perfect square...

  7. mukushla
    • 2 years ago
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    we have no 2 perfect square numbers with difference 2

  8. lgbasallote
    • 2 years ago
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    so i let m^2 + 2 = k^2

  9. lgbasallote
    • 2 years ago
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    @mukushla that's not really a proof.....

  10. shubhamsrg
    • 2 years ago
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    you may continue @igba

  11. lgbasallote
    • 2 years ago
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    hmm i suppose the next step would be m^2 - k^2 = 2 so (m+k)(m-k) = 2 i don't think this proves anything...

  12. mukushla
    • 2 years ago
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    actually that is :) \[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

  13. lgbasallote
    • 2 years ago
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    why so?

  14. shubhamsrg
    • 2 years ago
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    well,,you've almost proved it.. now m and k both are integers/.

  15. shubhamsrg
    • 2 years ago
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    and its k^2 -m^2 = 2 and not otherwise..

  16. lgbasallote
    • 2 years ago
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    oh yes

  17. lgbasallote
    • 2 years ago
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    2 = (k + m)(k-m)

  18. lgbasallote
    • 2 years ago
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    now...integers huh...

  19. mukushla
    • 2 years ago
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    \[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

  20. shubhamsrg
    • 2 years ago
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    you know k>m.. so from (k+m)(k-m)= 1*2 =>k+m =2 and k-m = 1 this doesnt have an integral solution..

  21. lgbasallote
    • 2 years ago
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    why do i know k > m again?

  22. shubhamsrg
    • 2 years ago
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    m^2 + 2 = k^2 both m and k are positive integers hence k>m

  23. shubhamsrg
    • 2 years ago
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    in layman language,, k^2 - m^2 >0

  24. lgbasallote
    • 2 years ago
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    when was it assumed that m and k are positive?

  25. shubhamsrg
    • 2 years ago
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    what else do you mean by perfect square ?

  26. lgbasallote
    • 2 years ago
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    i don't see how that relates.... the definition of a perfect square is x = k^2

  27. shubhamsrg
    • 2 years ago
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    4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..

  28. estudier
    • 2 years ago
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    @mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....

  29. lgbasallote
    • 2 years ago
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    (-2)^2 is also 4...

  30. shubhamsrg
    • 2 years ago
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    you take the absolute value

  31. shubhamsrg
    • 2 years ago
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    i mean you have to..

  32. lgbasallote
    • 2 years ago
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    why?

  33. shubhamsrg
    • 2 years ago
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    see,, (k+m)(k-m) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..

  34. lgbasallote
    • 2 years ago
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    ...i don't think this is a valid proof....

  35. shubhamsrg
    • 2 years ago
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    another explanation might me x is a perfect square because its y^2 x = y^2 = (-y)^2 sqrt(x) = | y | = | -y |

  36. jhonyy9
    • 2 years ago
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    who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???

  37. lgbasallote
    • 2 years ago
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    since when did sqrt x become |-y|

  38. estudier
    • 2 years ago
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    0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.

  39. jhonyy9
    • 2 years ago
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    so and i think that this difference of two will be the ,,key" of this proof of ...

  40. jhonyy9
    • 2 years ago
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    @lgbasallote what is your opinion from this ,please ?

  41. lgbasallote
    • 2 years ago
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    my opinion is that this is some tricky algebra

  42. jhonyy9
    • 2 years ago
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    how do you think it ???

  43. jhonyy9
    • 2 years ago
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    so and why ?

  44. lgbasallote
    • 2 years ago
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    because i can't think of a proof

  45. lgbasallote
    • 2 years ago
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    how can you prove/disprove that the difference of two squares is 2...

  46. lgbasallote
    • 2 years ago
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    hmmm...wait...

  47. jhonyy9
    • 2 years ago
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    so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much

  48. lgbasallote
    • 2 years ago
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    their fields of specialty aren't exactly in proving

  49. estudier
    • 2 years ago
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    What's wrong with my proof?

  50. lgbasallote
    • 2 years ago
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    anyway... m^2 + 2 = k^2 k^2 - m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2 - (2y^2) = 2 4x^2 - 4y^2 = 2 then... 2x^2 - 2y^2 = 1 x^2 - y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction

  51. lgbasallote
    • 2 years ago
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    and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

  52. lgbasallote
    • 2 years ago
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    @estudier i mean

  53. lgbasallote
    • 2 years ago
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    @mukushla too

  54. estudier
    • 2 years ago
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    So, that doesn't mean it's invalid..

  55. lgbasallote
    • 2 years ago
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    it does actually.....you can't do infinite substitutions to prove anything....

  56. estudier
    • 2 years ago
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    There is no infinity involved, all integer squares end in the numbers I gave.

  57. lgbasallote
    • 2 years ago
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    infinite loops involves infinity agree?

  58. estudier
    • 2 years ago
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    No infinite loop either, just logic.

  59. estudier
    • 2 years ago
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    An integer ends in one of digits 0 to 9. The squares must end in etc....

  60. lgbasallote
    • 2 years ago
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    but we both know that setting values is illegal in proving

  61. estudier
    • 2 years ago
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    I am not setting any values.

  62. lgbasallote
    • 2 years ago
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    and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

  63. lgbasallote
    • 2 years ago
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    you're checking the squares of each number

  64. estudier
    • 2 years ago
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    I am not.

  65. lgbasallote
    • 2 years ago
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    and though there is a pattern...there is still susbtitution involved

  66. lgbasallote
    • 2 years ago
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    and then you'll have to substitute forever to verify that the pattern really does not break

  67. lgbasallote
    • 2 years ago
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    then infinity is involved

  68. estudier
    • 2 years ago
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    It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

  69. lgbasallote
    • 2 years ago
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    so you do admit that you checked the 0 to 9 squares

  70. estudier
    • 2 years ago
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    Anything else is impossible.

  71. lgbasallote
    • 2 years ago
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    ...that's substitution

  72. lgbasallote
    • 2 years ago
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    anyway...my point is...substitution isn't really allowed in proving

  73. estudier
    • 2 years ago
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    You will find that this fact is used a lot in many valid proofs in number theory.

  74. lgbasallote
    • 2 years ago
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    perhaps

  75. estudier
    • 2 years ago
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    OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

  76. lgbasallote
    • 2 years ago
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    since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

  77. estudier
    • 2 years ago
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    I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)

  78. lgbasallote
    • 2 years ago
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    can you prove it? that it's a fact about squares?

  79. estudier
    • 2 years ago
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    Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

  80. lgbasallote
    • 2 years ago
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    you didn't really demonstrate anything...you just repeated what you said

  81. estudier
    • 2 years ago
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    You asked for proof, I just gave u a proof.

  82. estudier
    • 2 years ago
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    If you deny my proof, provide a counterexample.

  83. lgbasallote
    • 2 years ago
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    since you try so hard to defend your proof, i'll accept it

  84. estudier
    • 2 years ago
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    That's very gracious, thank you:-)

  85. lgbasallote
    • 2 years ago
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    do check on my algebra proof though please. i would like to know if that's the way contradiction works

  86. estudier
    • 2 years ago
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    OK, I will look at it now....

  87. estudier
    • 2 years ago
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    Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

  88. lgbasallote
    • 2 years ago
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    why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

  89. estudier
    • 2 years ago
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    No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

  90. lgbasallote
    • 2 years ago
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    oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

  91. lgbasallote
    • 2 years ago
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    again i ask... how is k > m

  92. estudier
    • 2 years ago
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    (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

  93. mukushla
    • 2 years ago
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    factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[k-m=1\]\[k+m=2\]add them\[2k=3\] Contradiction

  94. lgbasallote
    • 2 years ago
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    again i ask for the nth time @mukushla how is k > m

  95. mukushla
    • 2 years ago
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    ok man i just want to answer ur question

  96. estudier
    • 2 years ago
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    "2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?

  97. lgbasallote
    • 2 years ago
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    well because 8 - 4 is not 2 but it's even minus even

  98. lgbasallote
    • 2 years ago
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    @mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

  99. estudier
    • 2 years ago
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    OK, so finally, the algebra Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

  100. mukushla
    • 2 years ago
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    i thought they are positive :) sorry

  101. lgbasallote
    • 2 years ago
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    why (k+1)^2 @estudier ?

  102. estudier
    • 2 years ago
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    I already, said, that's the next square....

  103. lgbasallote
    • 2 years ago
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    next square?

  104. lgbasallote
    • 2 years ago
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    is this still related to my proof?

  105. estudier
    • 2 years ago
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    ? n is k^2, what is the next square?

  106. mukushla
    • 2 years ago
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    no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) ...now add them\[2|k|=3\]Contradiction and we are done

  107. estudier
    • 2 years ago
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    No, this proof Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

  108. lgbasallote
    • 2 years ago
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    are you allowed to do that? @mukushla

  109. lgbasallote
    • 2 years ago
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    can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

  110. mukushla
    • 2 years ago
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    why not...for any real number \[a^2=|a|^2\]

  111. estudier
    • 2 years ago
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    Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED

  112. estudier
    • 2 years ago
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    That's 2 proofs I have given now...

  113. lgbasallote
    • 2 years ago
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    a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?

  114. estudier
    • 2 years ago
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    Think about it....:-)

  115. lgbasallote
    • 2 years ago
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    in the meanwhile... @mukushla why |k| - |m| = 1?

  116. mukushla
    • 2 years ago
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    because\[|k|-|m|<|k|+|m|\]

  117. lgbasallote
    • 2 years ago
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    how does that make |k| - |m| = 1

  118. mukushla
    • 2 years ago
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    u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2

  119. mukushla
    • 2 years ago
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    right?

  120. lgbasallote
    • 2 years ago
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    who said those are positive?

  121. mukushla
    • 2 years ago
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    |k|>|m| right?

  122. lgbasallote
    • 2 years ago
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    something feels wrong...

  123. lgbasallote
    • 2 years ago
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    anyway...where did you get 2|k| > 3

  124. mukushla
    • 2 years ago
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    ok...think on what i said...because i believe thats the shortest way to prove ur statement

  125. lgbasallote
    • 2 years ago
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    |2k| = 3 *

  126. mukushla
    • 2 years ago
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    adding\[|k|-|m|=1\]and\[|k|+|m|=2\]

  127. estudier
    • 2 years ago
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    Going back to your question about 2, u can just adjust it to difference of 2q...

  128. lgbasallote
    • 2 years ago
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    im really skeptical about this... especially that |k|^2 - |m|^2 = 2

  129. lgbasallote
    • 2 years ago
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    difference of 2q?

  130. estudier
    • 2 years ago
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    You were worried about the validity of your statement, change it to 2q instead of just 2.

  131. estudier
    • 2 years ago
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    Or 2p, or whatever letter u like best.

  132. estudier
    • 2 years ago
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    1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)

  133. lgbasallote
    • 2 years ago
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    oh. when i said difference of two numbers equal to 2

  134. lgbasallote
    • 2 years ago
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    but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

  135. lgbasallote
    • 2 years ago
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    and @mukushla how did you get |k| > |m|

  136. estudier
    • 2 years ago
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    No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.

  137. lgbasallote
    • 2 years ago
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    and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?

  138. estudier
    • 2 years ago
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    Yes, the algebra is completely general...

  139. lgbasallote
    • 2 years ago
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    i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

  140. estudier
    • 2 years ago
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    ...which is greater than n+3, right?

  141. lgbasallote
    • 2 years ago
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    it is?

  142. estudier
    • 2 years ago
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    Number theorists don't pay attention to 0

  143. lgbasallote
    • 2 years ago
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    unless k is 0....

  144. lgbasallote
    • 2 years ago
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    why not?

  145. estudier
    • 2 years ago
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    Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.

  146. sauravshakya
    • 2 years ago
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    Can I try?

  147. lgbasallote
    • 2 years ago
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    ..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....

  148. estudier
    • 2 years ago
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    More the merrier.

  149. estudier
    • 2 years ago
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    "because i know nothing about it" But you do now:-)

  150. lgbasallote
    • 2 years ago
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    it's not like it's limited to the 99s....

  151. lgbasallote
    • 2 years ago
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    i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not

  152. estudier
    • 2 years ago
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    Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.

  153. sauravshakya
    • 2 years ago
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    Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2-m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.

  154. estudier
    • 2 years ago
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    Yes, that's similar to the one I gave...

  155. estudier
    • 2 years ago
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    "number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number

  156. estudier
    • 2 years ago
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    http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...

  157. jhonyy9
    • 2 years ago
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    ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?

  158. jhonyy9
    • 2 years ago
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    so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2 so m1,2 = +/- sqrt(x^2 -2)

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